# 1557. Minimum Number of Vertices to Reach All Nodes

## Description

Given a directed acyclic graph, with n vertices numbered from 0 to n-1, and an array edges where edges[i] = [fromi, toi] represents a directed edge from node fromi to node toi.

Find the smallest set of vertices from which all nodes in the graph are reachable. It's guaranteed that a unique solution exists.

Notice that you can return the vertices in any order.

Example 1:

Input: n = 6, edges = [[0,1],[0,2],[2,5],[3,4],[4,2]]
Output: [0,3]
Explanation: It's not possible to reach all the nodes from a single vertex. From 0 we can reach [0,1,2,5]. From 3 we can reach [3,4,2,5]. So we output [0,3].

Example 2:

Input: n = 5, edges = [[0,1],[2,1],[3,1],[1,4],[2,4]]
Output: [0,2,3]
Explanation: Notice that vertices 0, 3 and 2 are not reachable from any other node, so we must include them. Also any of these vertices can reach nodes 1 and 4.


Constraints:

• 2 <= n <= 10^5
• 1 <= edges.length <= min(10^5, n * (n - 1) / 2)
• edges[i].length == 2
• 0 <= fromi, toi < n
• All pairs (fromi, toi) are distinct.

## Solutions

• class Solution {
public List<Integer> findSmallestSetOfVertices(int n, List<List<Integer>> edges) {
var cnt = new int[n];
for (var e : edges) {
++cnt[e.get(1)];
}
List<Integer> ans = new ArrayList<>();
for (int i = 0; i < n; ++i) {
if (cnt[i] == 0) {
}
}
return ans;
}
}

• class Solution {
public:
vector<int> findSmallestSetOfVertices(int n, vector<vector<int>>& edges) {
vector<int> cnt(n);
for (auto& e : edges) {
++cnt[e[1]];
}
vector<int> ans;
for (int i = 0; i < n; ++i) {
if (cnt[i] == 0) {
ans.push_back(i);
}
}
return ans;
}
};

• class Solution:
def findSmallestSetOfVertices(self, n: int, edges: List[List[int]]) -> List[int]:
cnt = Counter(t for _, t in edges)
return [i for i in range(n) if cnt[i] == 0]


• func findSmallestSetOfVertices(n int, edges [][]int) (ans []int) {
cnt := make([]int, n)
for _, e := range edges {
cnt[e[1]]++
}
for i, c := range cnt {
if c == 0 {
ans = append(ans, i)
}
}
return
}

• function findSmallestSetOfVertices(n: number, edges: number[][]): number[] {
const cnt: number[] = new Array(n).fill(0);
for (const [_, t] of edges) {
cnt[t]++;
}
const ans: number[] = [];
for (let i = 0; i < n; ++i) {
if (cnt[i] === 0) {
ans.push(i);
}
}
return ans;
}


• impl Solution {
pub fn find_smallest_set_of_vertices(n: i32, edges: Vec<Vec<i32>>) -> Vec<i32> {
let mut arr = vec![true; n as usize];
edges.iter().for_each(|edge| {
arr[edge[1] as usize] = false;
});
arr.iter()
.enumerate()
.filter_map(|(i, &v)| if v { Some(i as i32) } else { None })
.collect()
}
}